Does anyone know how to parametrize the boundary of the Mandelbrot set? I am not a fractal-geometer or a dynamical systems person. I just have some idle curiosity about this question.

The Mandelbrot set is customarily defined as the set $M$ of all points $c\in\mathbb{C}$ such that the iterates of the function $z\mapsto z^2+c$, starting at $z=0$, remain bounded forever. Most very pretty depictions of the Mandelbrot set show $M$ as an intersection of an infinite sequence of sets $M_1\supset M_2\supset M_3\supset\cdots$, where the boundary of $M_i$ is the curve $|z_i(c)|=K$. Here $z_i(c)$ is the $i$th iterate of $z\mapsto z^2+c$, starting at $z=0$, and $K$ is some constant which guarantees that future iterates will escape. These curves $\partial (M_i)$ guide the viewer to see the increasingly intricate parts of the Mandelbrot set.

Each of these curves $\partial(M_i)$ is analytic and closed. They can thus be parametrized nicely with a trigonometric series. To be more specific, each boundary has a parametrization of the form
$$z(t)=\sum_{k=0}^\infty a_k\cos(kt)+i\sum_{k=0}^\infty b_k\sin(kt).$$
(In fact, since each boundary $\partial(M_i)$ is determined by a polynomial equation in the real and imaginary parts of $c$, I think each of these series should terminate. Correct me if I am wrong.) I would think that the limiting path should also have some nice parametrization with a trigonometric series. Is this limit the same for all $K$? If the limit is not the same for all $K$, then is there a limit as $K\rightarrow\infty$? What are the Fourier coefficients?

Your proposed boundary parametrization does not seem to be uniquely defined, since there is (as far as I know) no canonical unit-time parametrization, and the Fourier coefficients would be changed by a reparametrization.
–
S. Carnahan♦Dec 8 '10 at 15:47

6 Answers
6

Lasse's answer expanded: Let $\psi$ be the map of the exterior of the unit disk onto the exterior of the Mandelbrot set, with Laurent series
$$
\psi(w) = w + \sum_{n=0}^\infty b_n w^{-n} =
w - \frac{1}{2} + \frac{1}{8} w^{-1} - \frac{1}{4} w^{-2} +
\frac{15}{128} w^{-3} + 0 w^{-4} -\frac{47}{1024} w^{-5} + \dots
$$
Then of course the boundary of the Mandelbrot set is the image of the unit circle under this map. However, this depends on the (not yet proved) local connectedness of that boundary. Here, for the coefficients $b_n$ there is no known closed form, but they can be computed recursively. Of course we put $w = e^{i\theta}$ and then this is a Fourier series.

I am not quite sure what you are asking. The boundary of the Mandelbrot set certainly is not an analytic curve. In fact, a famous result of Shishikura shows that the boundary of the Mandelbrot set has Hausdorff dimension 2.

Indeed, it is not even known whether the boundary is a curve at all (i.e., locally connected): this is currently probably the most famous conjecture in one-dimensional holomorphic dynamics.

If the Mandelbrot set is locally connected, then there is a natural description of the boundary of the Mandelbrot set (as the boundary values of the Riemann map of the complement of $M$); this is also known to be a natural combinatorial description in many ways. However, as mentioned above, this parametrization is not analytic, or even $C^1$.

Lassse: I am asking about the boundary curves $\partial(M_i)$ which are analytic for all $i$ and all $K$. For example, if $K=2$, then $\partial(M_1)$ is the circle $|c|=2$, $\partial(M_2)$ is the curve $|c^2+c|=2$, $\partial(M_3)$ is the curve $|(c^2+c)^2+c|=2$, etc.
–
David RichterDec 8 '10 at 14:49

I thought you were asking about the limit of these curves, which is the boundary of the Mandelbrot set? I should mention that a more natural approximation of the boundary of the Mandelbrot set would be via the level sets of the uniformizing function of the complement of $M$ ("equipotentials"). If $K$ is sufficiently large, these equipotentials will be close to the curves you describe.
–
Lasse Rempe-GillenDec 8 '10 at 17:56

To expand on Gerald Edgar's answer, some key phrases for you to look into are "Douady-Hubbard potential" and "external rays." An external ray is the image of the ray $\arg z = \theta$ for fixed $\theta$ under Gerald's conformal map $\psi$. The Douady-Hubbard potential is just the harmonic conjugate of the external ray argument: it's the potential for which the external rays are the field lines.

I'm pretty sure it hasn't been proved that $\psi(\zeta)$ is well defined for all $\zeta$ on the unit circle, but I think it is conjectured to be so. (Sometimes this is phrased as saying that the external ray "lands.") However, the external rays at rational angles $2\pi m/n$ are known to land, and moreover, the dynamics at the landing points on the boundary is related to the fraction $m/n$ in a really nice way. (There's an analogy between the doubling map $\theta \mapsto 2\theta$ on the circle and holomorphic maps $z\mapsto z^2+c$, and the dynamics of $\theta$ under the first map are related to the dynamics of the $z\mapsto z^2+c_\theta$ map, where $c_\theta$ is the landing point of the corresponding ray on the boundary of the Mandelbrot set.) Thus, this parametrization of the boundary is indeed an important and natural object (if it is well defined, as conjectured).

My conjecture would be that such a parametrization would not work. Try something similar for a simpler (in certain viewpoints) structure such as a Koch snowflake. Would your approach to parametrization allow you to generate a function based on $n$, the number of recursive iterations used to generate the snowflake to a certain depth? I would think not. You might be able to, at least for the Koch curve, parametrize the "rubber band" hull around it, but that would be trivial for most recursively defined objects.

I would think that the boundary of the Mandelbrot set can not be parametrized. The Mandelbrot set is simply connected and has a Hausdorff dimension of two. I assume you mean parametrized by a single variable, but that would only work if the boundary had a a Hausdorff dimension of one.